The following data on x = average daily hotel room rate and y = amount spent on
ID: 3265573 • Letter: T
Question
The following data on x = average daily hotel room rate and y = amount spent on entertainment (The Wall Street Journal, August 18, 2011) lead to the estimated regression equation = 17.49 + 1.0334x. For these data SSE = 1541.4. b. Develop a 95% confidence interval for the mean amount spent on entertainment for all cities that have a daily room rate of $89 (to 2 decimals). C. The average room rate in Chicago is $128. Develop a 95% prediction interval for the amount spent on entertainment in Chicago (to 2 decimals).Explanation / Answer
Answer:
a). 95% CI when x=89 is (94.85, 124.08)
b). 95% PI when x=128 is (110.69, 188.84)
Regression Analysis
r²
0.740
n
9
r
0.860
k
1
Std. Error
14.839
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
4,378.5778
1
4,378.5778
19.88
.0029
Residual
1,541.4222
7
220.2032
Total
5,920.0000
8
Regression output
confidence interval
variables
coefficients
std. error
t (df=7)
p-value
95% lower
95% upper
Intercept
17.4915
24.8314
0.704
.5039
-41.2255
76.2084
x
1.0334
0.2318
4.459
.0029
0.4854
1.5814
Predicted values for: y
95% Confidence Intervals
95% Prediction Intervals
x
Predicted
lower
upper
lower
upper
Leverage
89
109.465
94.847
124.083
71.453
147.478
0.174
128
149.769
132.574
166.964
110.693
188.844
0.240
Regression Analysis
r²
0.740
n
9
r
0.860
k
1
Std. Error
14.839
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
4,378.5778
1
4,378.5778
19.88
.0029
Residual
1,541.4222
7
220.2032
Total
5,920.0000
8
Regression output
confidence interval
variables
coefficients
std. error
t (df=7)
p-value
95% lower
95% upper
Intercept
17.4915
24.8314
0.704
.5039
-41.2255
76.2084
x
1.0334
0.2318
4.459
.0029
0.4854
1.5814
Predicted values for: y
95% Confidence Intervals
95% Prediction Intervals
x
Predicted
lower
upper
lower
upper
Leverage
89
109.465
94.847
124.083
71.453
147.478
0.174
128
149.769
132.574
166.964
110.693
188.844
0.240
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